Help needed!!!!!

A person goes to sleep between 1am and 2 am and he wakes up when his watch shows such a time that the two hands interchange their respective places. He wakes up between 2am and 3am, how long does he sleep?

Consider only hour hand & minute hand.

#Algebra #HelpMe! #Clock

Easy Math Editor

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`2 \times 3`	$2 \times 3$
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Comments

The answer is $\boxed {\frac{12}{13}}$ hours.

The positions of hour-hand and minute-hand have a certain relationship. They are together at 00:00h or 12:00AM. Let 12:00AM position be $0^\circ$ , the time the person goes to sleep at $t_1$ and wakes up at $t_2$ , the angle made by the hour-hand at $t_1$ and minute-hand at $t_2$ be $\alpha$ and that by the hour-hand at $t_2$ and minute-hand at $t_1$ be $\beta$ . Note that the hour-hand makes $30^\circ$ on the dial in $1$ hour while the minute-hand makes $360^\circ$ in $1$ hour.

Therefore, we have $\quad 30t_1 = \alpha \space, \quad 360t_1 = 360 + \beta$ (as $t_1 > 1$ hour) $\quad \Rightarrow t_1 = \dfrac {\alpha}{30} = 1 + \dfrac {\beta}{360}$

Similarly, we have $\quad t_2 = \dfrac {\beta}{30} = 2 + \dfrac {\alpha}{360}$

The time that the person sleep, $t = t_2-t_1 = \dfrac {\beta-\alpha} {30} = 1 - \dfrac {\beta - \alpha}{360}$

$\dfrac {\beta-\alpha} {30} \left( 1 + \dfrac {1}{12} \right) = 1\quad \Rightarrow \dfrac {\beta-\alpha} {30} = t = \boxed{\frac {12}{13}}$ hours.

Chew-Seong Cheong - 6 years, 7 months ago

thanx a lot!!

Aneesh Kundu - 6 years, 7 months ago

Why you add 10 to y in the equation 10+y/60

surya Prabha - 4 years, 7 months ago

he slept at 1:10 and woke up at 2:05

Mehul Arora - 6 years, 7 months ago

According to the book the answer should be $55\dfrac{5}{13}\text{ mins}$

Aneesh Kundu - 6 years, 7 months ago

but the question says that consider minute and hour hand only

Mehul Arora - 6 years, 7 months ago

@Mehul Arora – yes but that doesn't mean that time can't be fractional

Aneesh Kundu - 6 years, 7 months ago

A similar question once appeared in INMO And I really think this is a gud one So pls reshare and write solutions

Aneesh Kundu - 6 years, 7 months ago

The book is correct. Let $x$ and $y$ be the minutes past $1 am$ and $2 am$ where the hands could be. Then we solve this system of equations, for start and end times, the difference being the time of sleeping:

$\dfrac { 10+y }{ 60 } =\dfrac { x }{ 5 }$

$\dfrac { 5+x }{ 60 } =\dfrac { y }{ 5 }$

We get $x=\dfrac { 125 }{ 143 }$ and $y=\dfrac { 70 }{ 143 }$ , from which we can work out the time of sleep

$(120+5+x)-(60+10+y)=\dfrac { 720 }{ 13 } =55+\dfrac { 5 }{ 13 }$

Michael Mendrin - 6 years, 7 months ago

Thanx a lot!!

Aneesh Kundu - 6 years, 7 months ago

Why u add 10 to y in the equation (10+y)/60

surya Prabha - 4 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$